Ion Short Report Template

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Carleton University *

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1001

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Chemistry

Date

Feb 20, 2024

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docx

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Uploaded by CaptainTitanium13383

Acid-Base Titration Name: Student number: Date: TA’s name: Lab section: Purpose Theoretically and experimentally determine the shape of acid-base titration curves and evaluate the effectiveness of antacid tablets using acid-base titration.

Procedure 1. Set up the Logger Pro: connect the pH sensor to the laptop and run the program. 2. Restart the pH sensor in the “calibration” section. Immerse the sensor in the buffers with pH 4 and stabilize it. The same procedure should be done with buffer 7. 3. Obtain 15 mL of 0.100 M HCl, and 60 mL of 0.100 M Standard NaOH solution. 4. Fill the burette with 50 ml of NaOH. 5. Pipet 10.00 mL of 0.100 M HCl into a clean 250 mL beaker and dilute it to about l00 mL with distilled water. 6. Add a magnetic stirring bar and 2 drops of a suitable indicator to the HCl solution and place the beaker on a magnetic stirrer. 7. Start the titration and collect the obtained pH and the amount of NaOH added when they are stabilized. For the instructions see the board 8. Continue adding NaOH until you have a full Acid Base plot, and your pH no longer changes significantly with the addition of NaOH. 9. Repeat steps 5 to 8 inclusive using 10.00 mL of 0.100 M acetic acid instead of HCl. 10. Transfer the powder quantitatively into a pre-weighed 250 mL beaker and re-weigh. 11. Add 100 mL of distilled water and stir magnetically. 12. Obtain ~50 mL of 0.500 M HCl. 13. Titrate the solution using the 0.500 M HCl. Add the first 2.0 mL of acid titrant in 0.5 mL increments and allow enough time between each addition for the pill to consume the acid clicking. 14. Continue titration, then as the pH begins to decrease more rapidly, use 1 mL increments to obtain a better-defined titration curve. Plot the titration curve. 15. Once the pH is 2.2 click “Stop”

Data: Table1. Titration of a strong acid (0.100 M HCl) by a strong base (0.100 M NaOH) Volume NaOH is added (mL) pH 0.00 1.15 2.00 1.82 4.00 1.92 6.00 2.12 8.00 2.53 9.00 3.02 9.20 3.17 9.40 4.83 9.60 5.30 10.00 9.28 10.20 10.95 10.40 11.14 10.60 11.25 10.80 11.37 11.00 11.41 12.00 11.67 14.00 11.94 16.00 12.08 18.00 12.20 20.00 12.28 The color has been changed when 10 ml of NaOH was titrated. Table2. Titration of a weak acid (0.100 M acetic acid) by a strong base (0.100 M NaOH) Volume NaOH is added (mL) pH 0.00 2.85 2.00 3.86 4.00 4.28 6.00 4.80 8.00 5.26

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9.00 5.67 9.20 5.76 9.40 5.86 9.60 5.90 10.00 6.05 10.20 6.45 10.40 6.73 10.60 7.75 10.80 9.50 11.00 9.81 12.00 10.33 14.00 11.37 16.00 11.39 18.00 11.59 20.00 11.70 The color has been changed when 10.8 ml of NaOH was titrated. Table3. Titration of a weak base (antacid tablet) by a strong acid (0.500 M HCl) Volume NaOH is added (mL) pH 0.00 7.20 0.05 6.25 1.00 6.00 1.50 5.76 2.00 5.59 4.00 5.21 6.00 4.99 8.00 4.83 9.00 4.79 10.00 4.75 11.00 4.73 12.00 4.71 14.00 4.60 16.00 4.51 18.00 4.38

19.00 4.28 20.00 3.33 21.00 2.28 22.00 1.74 23.00 1.55 The mass of the empty beaker= 105.56 The mass of the beaker with antacid =107.02 Antacid= (1.46+/-0.01)g Calculation: 1. Finding the strong acid (HCl) using the pH probe The steepest point (when the solution changes color) happened when 10 ml of NaOH was added to the strong acid, and at this point, the Ph was 9.28.  Find the number of moles of HCL (NaOH is 0.100 M) 𝑛 = 𝑐 × 𝑣 𝑛 = 0.100 mol/L × 0.0100 L n= 1.00 ∗ 10 − 3 mol Therefore, the number of moles of HCL is 1 ∗ 0 ∗ 10 − 3 mol.  Such as the ratio of NaOH and HCl is 1:1 we can say that HCL is 1.00 ∗ 10 − 3 mol.  The concentration of HCl: (volume of HCl is 10.ml ) C=n/v C= ¿ .00 ∗ 10 − 3 mol/0.01l= 0.1M Therefore, the concentration of HCl is 0.1M  The relative error of HCL concentration (the theoretical =0.*00M) Therelative error = Experimentalvalue − theorical value theorical value Therelative error = 0.1 − 0.1 0.1 Therelative error = 0 Therefore, the relative error is 0 which means that my result is identical to the theoretical.

2. Finding the strong acid (HCl) using the chemical indicator  Find the number of moles of NaOH added at a concentration of 0.100ml and the color change at the same concentration 𝑛 = 𝑐 × 𝑣 𝑛 = 0.1*00 mol/L × 0.0100 L n= 1.00 ∗ 10 − 3 mol  Find the number of moles of HCL HCL is 1.00 ∗ 10 − 3 mol.  The concentration of HCl: (volume of HCl is 10.ml ) C=n/v C= ¿ .00 ∗ 10 − 3 mol/0.01l= 0.1M Therefore, the concentration of HCl is 0.1M 3 . Finding the weak acid (acetic acid) using the pH probe. (10.8 ml of NaOH was drained before the solution reached the steepest point)  Find the number of moles of acetic acid (NaOH is 0.100 M) 𝑛 = 𝑐 × 𝑣 𝑛 = 0.100 mol/L × 0.0108 L n= ¿ .08 ∗ 10 − 3 mol Therefore, the number of moles of acetic acid is ¿ .08 ∗ 10 − 3 mol.  Such as the ratio of NaOH and Acetic Aced is 1:1 we can say that NaOH is ¿ .08 ∗ 10 − 3 mol.  The concentration of Acetic Acid: (volume of acetic acid is 10.ml ) C=n/v C= 1.08 ∗ 10 − 3 mol/0.01l= 0.*08M

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Therefore, the concentration of acetic acid is 0.108M  The relative error of Acetic Acid concentration (the theoretical =0.100M) Therelative error = Experimentalvalue − theorical value theorical value Therelative error = 0.108 − 0.1 0.1 Therelative error = 0 .08*100=* 4. Antacid Table Calculation [HCl] = 0.500 mol/L [stomach acid] = 0.100 mol/L Density ( 𝜌 ) of stomach acid = 1.00 g/mL  The volume of the second endpoint is equal to 10.20ml  Calculation of moles of HCl (The volume of stomach acid neutralized given the concentration of stomach acid is 0.100 mol/L and the concentration of the titrant (HCL) is 0.500 mol/L) n = cv n = 0.500 M ∗ 0.02 L n = 0.01 moles  The calculation of the volume (mL) of stomach acid V = 0.01 ¿ .1 V = 100 ml = 0.1L .  The weight of stomach acid is neutralized. m = v ∗ p m = 100 ml ∗ 1 g / ml m = 100 g  How many times its weight in stomach acid can the antacid neutralize. ¿ of × ¿ m ( stomachacid ) m ( antacid tablet ) ¿ of × ¿ 100 1.46 = 68.49 Therefore, the antacid tablet is able to neutralize stomach acid 68 times. Discussion:

Conclusion: The point where the strong acid (0.100 M HCL) and 0.100 M NaOH reach their endpoint occurs at the expected theoretical concentration of 0.100 M, resulting in a relative error of 0. However, in contrast to the strong acid, the experimental endpoint concentration of acetic acid and NaOH measures slightly higher at 0.108 M, leading to an 8% relative error. In the second part of the experiment involving an antacid tablet used to neutralize stomach acid, the results showed that 1.46 grams of the tablet can neutralize 68 times 100 ml of stomach acid.